Problem 18 - Algebra and Trigonometry

Question

\(11-22\) . Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$ x=\frac{1}{2} y^{2} $$

Step-by-Step Solution

Answer

Focus: \((\frac{1}{2}, 0)\), Directrix: \(x = -\frac{1}{2}\), Focal diameter: 2.

Step 1: Identify the Equation Form

Recognize that the given equation \( x = \frac{1}{2} y^2 \) represents a parabola in the standard form \( x = a(y-k)^2 + h \), where \((h,k)\) is the vertex. Here, the vertex \((h, k)\) is \((0, 0)\). Therefore, this parabola opens to the right, and the given equation is already in its simplest form.

Step 2: Find the Parameter 'a'

In the equation \( x = \frac{1}{2} y^2 \), compare it with the standard form \( x = a(y-k)^2 + h \). Here, \( a = \frac{1}{2} \), \( h = 0 \), and \( k = 0 \). The vertex is at the origin \((0,0)\).

Step 3: Calculate the Focus

For a parabola opening sideways described by \( x = a(y-k)^2 + h \), the distance from the vertex to the focus is \( \frac{1}{4a} \). Here \( a = \frac{1}{2} \), so \( \frac{1}{4a} = \frac{1}{4 \times \frac{1}{2}} = \frac{1}{2} \). Thus, the focus is at \( (\frac{1}{2}, 0) \).

Step 4: Determine the Directrix

The directrix of a parabola opening to the right is a vertical line to the left of the vertex. Its equation is \( x = h - \frac{1}{4a} \). With \( h = 0 \) and \( \frac{1}{4a} = \frac{1}{2} \), the directrix is \( x = -\frac{1}{2} \).

Step 5: Calculate the Focal Diameter

The focal diameter (or latus rectum length) of a parabola is given by \( \frac{1}{|a|} \). With \( a = \frac{1}{2} \), the focal diameter is \( \frac{1}{\frac{1}{2}} = 2 \).

Step 6: Sketch the Graph

Plot the vertex at \((0, 0)\), the focus at \((\frac{1}{2}, 0)\), and draw the directrix as the line \( x = -\frac{1}{2} \). Illustrate the parabola opening to the right. The curve will pass through points that are equidistant from the focus and directrix.

Key Concepts

Focus of a Parabola

The focus of a parabola is a significant point that helps determine its shape and position. Imagine a parabola as a curve where every point is equidistant from the focus and another fixed line called the directrix.
In mathematical terms, the focus is a point located inside the parabola. For a sideways opening parabola of the form \( x = a(y-k)^2 + h \), the focus can be found using the formula:

• The distance from the vertex to the focus is \( \frac{1}{4a} \).
• The coordinates of the focus are \((h + \frac{1}{4a}, k)\).

In our exercise, the parabola equation is \( x = \frac{1}{2} y^2 \), where the vertex is at the origin \((0, 0)\) and \(a = \frac{1}{2}\). Thus, the focus is found at \((\frac{1}{2}, 0)\). This means the parabola will extend outward from this point, curving outward to the right. Knowing the focus helps in graphing the parabola accurately.

Directrix of a Parabola

The directrix of a parabola serves as a reference line equidistant from all points of the curve to the focus. In simple terms, the directrix guides how far the parabola stretches along its vertical or horizontal axis. It plays an important role in defining the parabolic shape.
For parabolas like the ones described by \( x = a(y-k)^2 + h \), the directrix is a vertical line for horizontal-opening parabolas. Its equation can be written as:

• \(x = h - \frac{1}{4a}\)

Understanding this, if we substitute values from our equation \( x = \frac{1}{2} y^2 \) into the formula, the vertex \(h = 0\) and \(\frac{1}{4a} = \frac{1}{2}\), then the directrix is located at \(x = -\frac{1}{2}\). With the directrix in place, one can visualize the parabola's balance and symmetry between the focus and the curve of the parabola itself.

Focal Diameter of a Parabola

The focal diameter, also known as the latus rectum, provides insight into the width of the parabola. Specifically, it measures how wide the parabola is through its focal point. This line segment is perpendicular to the axis of the parabola and passes through the focus.
To find the focal diameter for a parabola in the form \( x = a(y-k)^2 + h \), the formula to use is:

• \(\frac{1}{|a|}\)

For the equation \( x = \frac{1}{2} y^2 \), the value of \(a\) is \(\frac{1}{2}\). Therefore, the focal diameter is \(\frac{1}{\frac{1}{2}} = 2\). This indicates that the width of the parabola at the focus is 2 units long.
The focal diameter is essential for accurately sketching the curve, showing how the parabola opens wide from its vertex through the focus.